Question

Use the information given about the angle $0<\theta <2\mathrm{\pi }$, to find the exact value of

$\cos \theta =\frac{-\sqrt{6}}{3}$

Short answer

Part (a). $\sin 2\theta =-\frac{2\sqrt{2}}{3}$

part (b). $\cos 2\theta =\frac{1}{3}$

part (c).$\sin \frac{\theta }{2}=\sqrt{\frac{3+\sqrt{6}}{6}}$

part(d). $\cos \frac{\theta }{2}=\sqrt{\frac{3-\sqrt{6}}{6}}$

part (e). $\tan 2\theta =-2\sqrt{2}$

part (f).$\tan \frac{\theta }{2}=\sqrt{\frac{3-\sqrt{6}}{3+\sqrt{6}}}$

Step-by-step solution

Part (a) Step 1. Value of sin2θ

Angle is in the second quadrant.

So, sine is positive and cosine is negative.

So,

$$\begin{gathered} \sin \theta =\sqrt{1-{\cos }^2\theta } \\ \sin \theta =\sqrt{1-{\left(-\frac{\sqrt{6}}{3}\right)}^2} \\ \sin \theta =\sqrt{1-\frac{6}{9}} \\ \sin \theta =\frac{1}{\sqrt{3}} \end{gathered}$$

We know that,

$$\begin{gathered} \sin 2\theta =2\sin \theta \cos \theta \\ \sin 2\theta =2\times \frac{1}{\sqrt{3}}\times \frac{-\sqrt{6}}{3} \\ \sin 2\theta =-\frac{2\sqrt{2}}{3} \end{gathered}$$

Part (b) Step 1. value of cos2θ

We know that,

$$\begin{gathered} \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \\ ={\left(\frac{-\sqrt{6}}{3}\right)}^2-{\left(\frac{1}{\sqrt{3}}\right)}^2 \\ =\frac{6}{9}-\frac{1}{3} \\ =\frac{1}{3} \end{gathered}$$

part (c) Step 1. Value of sinθ2

We know that,

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ =\sqrt{\frac{1-\left(-{\displaystyle \frac{\sqrt{6}}{3}}\right)}{2}} \\ =\sqrt{\frac{3+\sqrt{6}}{6}} \end{gathered}$$

Part (d) step 4. Value of cosθ2

We know that,

$$\begin{gathered} \cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}} \\ =\frac{\sqrt{1+\left(-{\displaystyle \frac{\sqrt{6}}{3}}\right)}}{2} \\ =\sqrt{\frac{3-\sqrt{6}}{6}} \end{gathered}$$

Part (e) step 1. Value of  tan2θ

We have:

$$\begin{gathered} \sin 2\theta =-\frac{2\sqrt{2}}{3} \\ \cos 2\theta =\frac{1}{3} \end{gathered}$$

We know that,

role="math" localid="1646561005169" $$\begin{gathered} \tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta } \\ =\frac{-{\displaystyle \frac{2\sqrt{2}}{3}}}{{\displaystyle \frac{1}{3}}} \\ =-2\sqrt{2} \end{gathered}$$

Part (f) step 1. value of tanθ2

We have:

$$\begin{gathered} \sin \frac{\theta }{2}=\sqrt{\frac{3+\sqrt{6}}{6}} \\ \cos \frac{\theta }{2}=\sqrt{\frac{3-\sqrt{6}}{6}} \end{gathered}$$

We know that,

$$\begin{gathered} \tan \frac{\theta }{2}=\frac{\sqrt{{\displaystyle \frac{3+\sqrt{6}}{6}}}}{\sqrt{{\displaystyle \frac{3-\sqrt{6}}{6}}}} \\ =\sqrt{\frac{3-\sqrt{6}}{3+\sqrt{6}}} \end{gathered}$$